The roots of equation (q - r)x 2 + (r - p)x + (p - q) = 0 are equal. Prove that: 2q = p + r, that is, p, q & r are in A.P.

Given,

The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (r - p)² - 4 × (q - r) × (p - q) = 0

⇒ r² + p² - 2pr - 4(qp - q² - rp + qr) = 0

⇒ r² + p² - 2pr - 4qp + 4q² + 4rp - 4qr = 0

⇒ r² + p² + 2pr - 4qp - 4qr + 4q² = 0

⇒ (r + p)² - 4q(p + r) + 4q² = 0

Let p + r = y,

⇒ y² - 4qy + 4q² = 0

⇒ (y - 2q)² = 0

⇒ y - 2q = 0

⇒ y = 2q

⇒ p + r = 2q.

Hence, proved that p + r = 2q.